Answer:
9.4 m
Explanation:
We can use a moving frame of reference with the same speed as the car. From this frame of reference the car doesn't move. The origin is at the back of the car, the positive X axis points back and the positive Y axis points up.
If the ballon is launched at 9.7 m/s at 39 degrees of elevation.
Vx0 = 9.7 * cos(39) = 7.5 m/s
Vy0 = 9.7 * sin(39) = 6.1 m/s
If we ignore air drag, the baloon will be subject only to the acceleration of gravity. We can use the equation of position under constant acceleration.
Y(t) = Y0 + Vy0 * t + 1/2 * a * t^2
Y0 = 0
a = -9.81 m/s^2
It will fall when Y(t) = 0
0 = 6.1 * t - 4.9 * t^2
0 = t * (6.1 - 4.9 * t)
t1 = 0 (this is when the balloon was launched)
0 = 6.1 - 4.9 * t2
4.9 * t2 = 6.1
t2 = 6.1 / 4.9 = 1.25 s
The distance from the car will be the horizonta distance it travelled in that time
X(t) = X0 + Vx0 * t
X(1.25) = 7.5 * 1.25 = 9.4 m
The answer is A because you have to have erosion
Answer:
The coefficient of static friction between the box and floor is, μ = 0.061
Explanation:
Given data,
The mass of the box, m = 50 kg
The force exerted by the person, F = 50 N
The time period of motion, t = 10 s
The frictional force acting on the box, f = 30 N
The normal force on the box, η = mg
= 50 x 9.8
= 490 N
The coefficient of friction,
μ = f/ η
= 30 / 490
= 0.061
Hence, the coefficient of static friction between the box and floor is, μ = 0.061
Answer:
i) E = 269 [MJ] ii)v = 116 [m/s]
Explanation:
This is a problem that encompasses the work and principle of energy conservation.
In this way, we establish the equation for the principle of conservation and energy.
i)
![W_{1-2}= (F*d) - (m*g*h)\\W_{1-2}=(500000*2.5*10^3)-(40000*9.81*2.5*10^3)\\W_{1-2}= 269*10^6[J] or 269 [MJ]](https://tex.z-dn.net/?f=W_%7B1-2%7D%3D%20%28F%2Ad%29%20-%20%28m%2Ag%2Ah%29%5C%5CW_%7B1-2%7D%3D%28500000%2A2.5%2A10%5E3%29-%2840000%2A9.81%2A2.5%2A10%5E3%29%5C%5CW_%7B1-2%7D%3D%20269%2A10%5E6%5BJ%5D%20or%20269%20%5BMJ%5D)
At that point the speed 1 is equal to zero, since the maximum height achieved was 2.5 [km]. So this calculated work corresponds to the energy of the rocket.
Er = 269*10^6[J]
ii ) With the energy calculated at the previous point, we can calculate the speed developed.
![E_{k2}=0.5*m*v^2\\269*10^6=0.5*40000*v^2\\v=\sqrt{\frac{269*10^6}{0.5*40000} }\\ v=116[m/s]](https://tex.z-dn.net/?f=E_%7Bk2%7D%3D0.5%2Am%2Av%5E2%5C%5C269%2A10%5E6%3D0.5%2A40000%2Av%5E2%5C%5Cv%3D%5Csqrt%7B%5Cfrac%7B269%2A10%5E6%7D%7B0.5%2A40000%7D%20%7D%5C%5C%20v%3D116%5Bm%2Fs%5D)
